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18:01
@anakhro What you think? :-)
18:14
@Wolgwang We don't? If A is a set, A is in the powerset of A, always.
Watch out. $\{\emptyset\}$ is different from $\emptyset$.
@MikeMiller I see that Thorgott gives a counterexample to just substituting in x-a.
No, @Thor is doing a different function.
DETAILS matter!
@anakhro O_o
So Power set of $\{\phi\}$ is $\{\phi,\{\phi\}\}$?
Is this any different from considering $\{a\}$?
18:19
Yes...
What is the power set of $\emptyset$?
@TedShifrin But it's still a counterexample to just substituting in x-a into a power series about zero to get one around a.
But that is not what you're talking about.
@TedShifrin $\{\phi\}$
Of course. $z= a+(z-a)$, so if $f(z)=z$, then $f(z)=a+(z-a)$.
OK, agreed, @Wolgwang.
18:23
Maybe I am not sure what I am talking about at this point, Ted. I am trying to understand what is written in a book and I am grasping at straws at where they are getting this Laurent expansion around $a$ from (when at this point, the only thing we are given is the expansion for around $z=0$).
ahh, stop using phi for the empty set
use \emptyset
Thorgott is first to break.
No, I just used the correct symbol :P
They're not looking at the Laurent expansion of $\wp(z)$, @anakhro. They're looking at the Laurent expansion of $\wp(z-a)$. This is why I said you have to think about the composition of functions.
IMHO, you're making a mountain out of a speck of dust.
What's the name for an element in an abelian group that can not be written as a proper multiple of another element in the group?
18:26
Yes, I probably am. I just feel so out of my element with this elliptic function stuff.
It's been a frustrating week trying to understand some of this stuff.
Just think about $f(x)=x^2$. Then of course $f(x-a) = (x-a)^2$. This is different from writing $f(x)=(a+(x-a))^2 = a^2 + 2a(x-a) + (x-a)^2$.
We're talking definition of a function. NOTHING to do with elliptic functions.
I have no idea of the context. I'm just paying attention to the literal text you quoted.
I think I see what you mean. Because an expansion around z near a is f(z) = ... rather than expansion of f(z-a) (near zero) which is f(z-a) = ...
Roughly the right idea?
Right.
Different functions.
The issue is at the level of precalculus, not at the level of graduate complex analysis.
So that's what Mike meant about right answer to the wrong question. The substitution thing is the right answer to the question of finding the expansion of f(z-a), but not f(z) near a.
Well, if you do the latter, you have to do the algebra. I just did two examples for you.
18:32
Okay, now let me see if I can understand what the book says. Thanks for being patient and helping me, all.
@Thorgott I think the term is primitive
@Wolgwang I still don't know which one is right :-/
@Wolgwang what does it mean for B to be a subset of A?
@anakhro It means that all the elements of B belong to A.
Great, so let's test to see if your teacher is wrong. You claim that $\{\emptyset\}$ is not a subset of $\{\emptyset,1\}$. What are the elements of the former?
18:39
Pink elephant
Well when you want to try, feel free to ping me.
@anakhro I had read it somewhere sorry :-(
Hint: Every element of the empty set is a pink elephant. Or an element of $X.$ (No joke) — Dan Christensen Jan 30 '14 at 4:59
so let's say I got this limit and it doesn't commute with a colimit I like. Let's say the category is nice enough for us to do interesting things (probably we'll need it to be Abelian for reasons I can't remember). How do I figure out "how bad" they don't commute and is there a way to do it in general?
maybe I haven't given enough details
locks @BigSocks in the category room with @Thor
I am finally safe
18:46
Somehow I think your textbook is a clearer source than a 0-upvote comment on a 7 year old thread.
Was that supposed to make sense to the casual observer?
He's making a joke which you will understand when you understand the empty set ... the point is that there is no such thing as a pink elephant ...
Ohhhh ... that.
My elephants usually arrive purple.
@anakhro No element is present in it...
@Wolgwang are you sure? What are the elements of {x}?
18:49
@TedShifrin with yellow flowers
@anakhro x only
Great. Now let $x=\emptyset$ and what do you notice?
@robjohn Soon we should be at e e cummings.
Element of {x} is $\emptyset$ ..
This is correct. So $\emptyset$ has no elements, but $\{\emptyset\}$ has one element.
18:55
A box containing an empty box is not an empty box
And so does $\{\emptyset,\emptyset\}$
:-o
Thanks :-)
@TedShifrin do you like ee?
Why if $\lambda$ is in the spectrum of some continuous linear transformation $T$ then $\lambda ^n$ is in the spectrum of $T^n$?
Start by spelling out what the words mean...
19:00
I know that $T-\lambda^n I$ is of the form $T-\lambda I$ times something
Mike, I don't assume finite dimensional
So it's not clear to me why (non-invertible) times (something) cannot be invertible
Yes, this was clear from the word continuous. It still fails to be surjective or injective, you just now have to do a case analysis instead of focusing on kernel as in the f.d.case. Maybe what you're missing is that $B = T^{n-1} + \lambda T^{n-2} + \cdots + \lambda^{n-1} I$ has $B(T - \lambda I) = (T - \lambda I)B = T^n - \lambda^n I$.
I am not happy with this because in reality I need it in a general Banach algebra
The book claims that if $\lambda \in \sigma (x)$ then $\lambda ^n \in \sigma (x^n)$
I agree that $B(x-\lambda I)=x^n-\lambda ^n I$
Suppose you randomly pick a collection of balls on the Poincare disk $\Bbb H^2$.
Choose a point not in this collection. Would you be able to see the ideal boundary from that point?
If you don't ask the question you really mean then obviously you aren't going to get the answer you want. I don't immediately know the answer to the Banach algebra question, but I suspect you just want to argue that $(ab)^{-1}$ commutes with both $a$ and $b$ separately when $ab$ is invertible and $ab = ba$.
From which it will follow that both $a$ and $b$ are invertible with inverses $b(ab)^{-1}$ and $(ab)^{-1} a$ respectively.
Could be the wrong approach though, I don't know yet that this is true.
i.e., can random obstacles in your view clog the sky?
19:16
I presume the collection is infinite?
@TedShifrin Yeah
You don't need that much, it's sufficient to argue a has a right and left inverse whenever ab has a right and left inverse and a and b commute
I think I've run into something like this before, a @Balarka, many years ago. Cool question.
Yeah but can you argue that
I'm stumbling
That's interesting! It occurs to me I need to say something like this if I want to do percolation in negative curvature spaces
19:18
ab invertible means there is a c s.t. abc = 1 = cab = cba, so cb anc bc are respectively left and right inverses
Oh of course, the point is that if an element in a ring has a left and right inverse those elements must coincide
Which is in fact how you'd prove my commutativity claim
right
0
Q: looking for $3D$ vector field satisfying certain projection conditions

geocalc33I'm searching for a $3D$ vector field $V$ in $(0,1)^3$ whose parallel projections onto the boundary of $[0,1]^3$ are the following $2D$ vector fields: The parallel projection of $V$ onto the $x-y$ plane will yield the vector field $\{x\log(x),-y \log(y)\}.$ The parallel projection of $V$ onto the...

I'm open to hints!
@Balarka: So if the countable collection of balls is dense, real trouble.
@Emolga See the above discussion. This is pure algebra, no analysis. If a,b are elements in a monoid so that ab is invertible and ab = ba --- meaning (ab)c = c(ab) = 1 for some element c --- then in fact a, b are invertible. To see this, observe that a has a right inverse and a left inverse: a(bc) = 1 and c(ab) = c(ba) = (cb)a = 1. Now observe that because cb = (cb)(abc) = (cba)(bc) = bc in fact the two inverses coincide and a is invertible. Similarly b.
19:22
That should be a 0 probability event
Notice the essential use of commutativity, which is why this doesn't reflect examples like the shift operators where a,b can fail to be invertible even though ab is.
You also need to argue that B is a bounded operator
@Astyx @MikeMiller I understand now, thank you very much!!
????? In our explicit case $b = a^{n-1} + \lambda a^{n-2} + \cdots + \lambda^{n-1}$
no argument needed...
@Emolga For sure. Notice that the POV one uses when thinking about B(H) is quite different than the POV one uses when thinking about an abstract Banach algebra :)
if you're working with C* algebras by Gelfand-Naimark the two POV's coincide but that's using heavy machinery
What's an analogue of Stone-Weierstrass theorem for topological Banach algebras?
19:26
Yes, I now realize that
There must be one, right
TIL Naimark's name is "Mark Naimark"
Troll parents
lol didn't know that either
@MikeMiller Yeah sure, but you still need to keep that in mind
If ab is invertible but b is not bounded, then you can't say anything about a
Sure but nobody ever said any sentence in a level of generality which suggested b was an unbounded operator :) That would be unnatural
19:29
Astyx is actually German, not French
I get your point but I think it's pedantry
@BalarkaSen They're the same, mathematically
Thurston school bro
Pierre Pansu, Francois Laudenbach
et al
Perhaps it is :)
That's uhhhh uhhhhhh the exception that proves the rule
apparently Thurston was an inspiration for Grothendieck. What exactly he inspired Grothendieck with I do not know
I just know it is true
19:32
Wow. I thought Grothendieck was mostly done by the time Thurston came along.
Yeah he was thinking about Galois-Teichmuller Yoga lmao
that must be the inspitation then
Maybe he was inspired to stop writing down his results
loool
good one
ok yeah it is GT
(one of the authors, Athanese Papadopoulos, is from the Thurston school - i forgot to mention him above)
Most greek name I have ever heard
yeah but i think he was born and brought up in the french school mathematically
i could be wrong
19:38
he has huge 3 volume books on Teichmuller theory
I saw them in the library
yeah student of Laudenbach
Just amused
i always think of the tintin character when i hear papadopoulos
oh lol rastapopoulos is also greek
Rastafarian !
are there any good math podcasts out there that everyone can recommend?
19:41
I think Papakyriakopoulos whenever I hear papa
LOL
impossible name
had to edit twice
I remember that!
greatest proof of all time in basic topology
"Papa's theorems"
19:42
@geocalc33 honestly there should be
I struggled for a long time to understand that for my reading course before getting started on research
i straight up did not understand it for ~3 years
i think i understood it very recently.
for a long time i wanted to modify the self-intersecting disk to make it an embedded disk
It's very clever! It's one of these things that makes no sense until you understand it when it is very clear.
papa avoids it completely
he just takes covers and just gets a new guy in the cover
philosophically it is like unwinding but thats not how you say the proof at all
I distinctly remember being asked to explain this and giving some handwaving thing because I didn't really understand it. I think my explanation only makes sense if the tendril goes around once.
19:45
hahah
<--- definitely NO topologist
Very trial by fire summer
speaking of summer, I hear things happened in Texas
Futurism finally?
Opposite of summer.
19:46
Learnt 3-manifolds, kirby calculus for 4-manifolds, symplectic topology, knot theory, a little contact topology
good, wish i could read that many things at the same time
i read so little
I know none of that.
there should be undergraduate courses on 3manifold topology because otherwise no one will learn it
thats the point of undergraduate courses, teach dead subjects
Hard to do that without algebraic topology first.
yeah, that's annoying
19:49
I aim to please (being annoying, that is).
It was fun but those were 80 hour weeks. It's not something I could ever do again.
this was your 1st year grad school i assume?
found a pretty good episode of a podcast
It's called "Oxford Mathematics Public Lecture: Henry Segerman"
Summer after first year
Cool
All I have been reading is probability. Feels bad man
19:58
One day I'll take a year off and just read combinatorics for a year with no human contact
why "no human contact"?
I think that comes automatically when you do combinatorics
What nonsense
@BalarkaSen I'll catch up on my human contact the next year after I truly understand combinatorics
is understanding combinatorics even a thing
20:03
Yes
Though I do think we should rename all intro courses to "Counting 101"
Count the number of self-avoiding walks of length $n$ in $\Bbb Z^2$
Or $\Bbb Z^d$ in general
If you can do it you get Fields medal
should we also rename intro topology courses to "Wobble 101"
This is where they do it for the honeycomb lattice
@Thorgott No, call them Introduction to Rubber
also good
the continuation will be An Introduction to Glue
20:08
[0,1] is a quotient of the Cantor set? more like you can glue a cloud of particles together into a line
obvious
Just glue together the endpoints of every removed interval
@BalarkaSen Get me a teaching free job at ISI and we will do it
Why do they say "don't work with the square lattice self avoiding walk?"
@MikeMiller Too hard, let's rob a bank instead
We split the money. You don't have to teach and I don't have to do grad school
20:16
Well now that you've posted about it the jig is up
They don't know that we know that they know
takes notes furiously
Yeah now I am racing to solve this problem
If I were a random walk how would I move?
randomly walk towards a bank and rob it
Gambler's ruin
but its the bank
20:21
Solid bit Thor
hahaha
1
Q: Parametrization of curve in $\Bbb R^3$

34PancakesI’m trying to find the parametrization of a curve in $\Bbb R^3$ satisfying $p_1=(0,1,1)$ and $p_2=(1,0,0)$ that parallel projects to the x-y plane onto $\log(x)\log(y)=1.$ I think that maybe I can do a rotation on the x-y plane curve and then scale its length to satisfy the endpoint conditions? A...

can someone upvote so they can talk in chat?
Hi @user889830
am I understanding correctly that the total spaces of S^3-bundles over S^4 still haven't been fully classified up to difeomorphism?
20:35
I try not to get the phrases "Rob the bank I'm at" vs "I'm Rob at the bank" mixed up.
@Thorgott Not a bed time story for horses...
Every classification you could possibly want
The diffeomorphism classification is in Corollary 1.6 and has you reduce mod 56n. :)
the Crowley-Escher paper doesn't have a complete diffeomorphism classification, but it seems the Crowley-Nordström paper may have it
Oh, you aren't satisfied with oriented diffeomorphism?
1.6 only tells when $M_{m,n}$ and $M_{m^{\prime},n}$ are diffeomorphic, but it doesn't seem to say anything about $M_{m,n}$ versus $M_{m^{\prime},n^{\prime}}$
The n can be read off from H_4 of the total space
Or at least |n|
H_4(M_{m,n}) = Z/n
20:45
duh, I forgot about that
thanks
And then for orientation-reversal, the orientation-reversal of M_{m,n} is M_{-m-n,n} from Rmk 1.1
It's not so obvious to me how the cases of n and -n relate admittedly
Here are some combinatorics questions to chew on if you want
p is prime.

(1) Give a combinatorial proof that C(pa, pb) = C(a,b) mod p (This is easy, I had a proof in a few minutes)

(2) Give a combinatorial proof that C(pa,pb) = C(a,b) mod p^2. (Still working on this)

(3) p >= 5. Give a combinatorial proof that C(pa, pb) = C(a,b) mod p^3. (I think this is open)
but you just said how they relate?
Oh I see I should have read their remark more carefully. Notice that my n is unchanged there. But their remark implies that $M_{m,n} \cong M_{m+n,-n} \cong M_{-m, -n} \cong M_{-m-n, n}$ unorientedly
Whaat I wrote above is an oriented diffeomorphism I think
Oh I've been foolish it literally says the orientation reversing case right there, and then these formulas tell you how to see if M_{m,n} and M_{k, -n} are oriented diffeo or not --- check whether M_{m,n} and M_{-k, n} are or-reversing diffeo using Cor 1.6
@BalarkaSen He turned the bank inside-out?

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